leetcode题解Java | 310. Minimum Height Trees

题目：https://leetcode.com/problems/minimum-height-trees/#/description

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function
to find all the MHTs and return a list of their root labels.

Format

The graph contains 

n

 nodes which are labeled from 

0

 to 

n
- 1

. You will be given the number 

n

 and a list of undirected 

edges

 (each
edge is a pair of labels).

You can assume that no duplicate edges will appear in 

edges

. Since all edges are undirected, 

[0,
1]

 is the same as 

[1, 0]

 and thus will not appear together in 

edges

.

Example 1:

Given 

n = 4

, 

edges
= [[1, 0], [1, 2], [1, 3]]

0
|
1
/ \
2   3

return 

[1]

Example 2:

Given 

n = 6

, 

edges
= [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

0  1  2
\ | /
3
|
4
|
5

return 

[3, 4]

Hint:
How many MHTs can a graph have at most?

Note:

(1) According to the definition of tree on Wikipedia: “a tree is an undirected graph
in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

分析：

解法一：用类型拓扑排序的方法，每次删除度为一的节点，最后剩下的（1个或2个节点），就是root

解法二：找最长路径，最长路径的中点就是root。而最长路径的找法是，先DFS，最长路的末端为起点，再做DFS。注意：递归的时候，不要把大数组放进去，容易栈溢出

Java实现：

public class Solution {
List<Integer>[] lists;
public List<Integer> findMinHeightTrees(int n, int[][] edges)
{
lists = new ArrayList
;
for(int i=0; i<n; ++i)
lists[i] = new ArrayList<>();
for(int i=0; i<edges.length; ++i)
{
int v1 = edges[i][0];
int v2 = edges[i][1];
lists[v1].add(v2);
lists[v2].add(v1);
}
List<Integer> ans =
9c7a
new ArrayList<>();
List<Integer> maxList = DFS(0, n, -1);
maxList = DFS(maxList.get(0), n, -1);
int len = maxList.size();
if(len%2==1)
ans.add(maxList.get(len/2));
else
{
ans.add(maxList.get(len/2));
ans.add(maxList.get(len/2-1));
}
return ans;
}
public List<Integer> DFS(int k, int n, int pre)
{
List<Integer> maxList = new ArrayList<>();
for(int i=0; i<lists[k].size(); ++i)
{
int v = lists[k].get(i);
if(v!=pre)
{
List<Integer> tmp = DFS(v, n, k);
if(tmp.size()>maxList.size())
maxList = tmp;
}
}
maxList.add(k);
return maxList;
}
}